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I need help on solving inequalities?

  1. Oct 27, 2005 #1
    I need help on solving inequalities? Someone please help me. I 'm currently taking Pre-calculus. :smile:
     
  2. jcsd
  3. Oct 27, 2005 #2
    Like

    [tex]x^2+2x-1\geq0[/tex]
    ?
     
  4. Oct 27, 2005 #3
    equations like the one implied ,
     
  5. Oct 27, 2005 #4
    Ok,
    This is the way described through my math teachers, although it is not exactly the method I use now.
    [tex]x^2+2x-3\geq[/tex]
    First, solve the problem as if it were a normal equality.
    [tex]x^2+2x-3=0[/tex]
    and you get
    [tex]x_1=1 x_2=-3[/tex]
    Now, write down all sets of numbers between those.
    [tex](-\infty,-3]
    [-3,1]
    [1,\infty)[/tex]
    Now set up a table like so
    .....set.....|||||||||||sample point|||||||||||(x-1)|||||||||||(x+3)|||||||||||+ or - ?
    [tex](-\infty,-3][/tex]....|||||||||||....-4.....|||||||||||..-..|||||||||||..-..|||||||||||....+...
    [tex][-3,1][/tex].......|||||||||||.....0......|||||||||||..-..|||||||||||..+..|||||||||||....-...
    [tex][1,\infty)[/tex].......|||||||||||.....2......|||||||||||..+..|||||||||||..+..|||||||||||....+...


    Now, since it was greater than or equal to, you know it has to be greater than zero, therefore the positive ones are the ones you want.

    Therefore, the two sets [tex](-\infty,-3][/tex] and [tex][1,\infty)[/tex] work.


    Now, you know that your answer is [tex](-\infty,-3]\cup[1,\infty)[/tex]


    now, try to solve this one on your own

    [tex]x^2-5x+6\geq0[/tex]
     
    Last edited: Oct 27, 2005
  6. Oct 27, 2005 #5
    You can do the familiar algebraic manipulation with inequalities, provided you remember to reverse the direction of the inequality whenever you multiply (or divide) by a negative quantity and practice simple logic. So by the example above,
    [tex]x^2+2x-3 \geq 0[/tex]
    [tex](x-1)(x+3) \geq 0[/tex]
    Now, if [itex]ab \geq 0[/itex], either ([itex]a \geq 0[/itex] and [itex]b \geq 0[/itex]) or ([itex]a \leq 0[/itex] and [itex]b \leq 0[/itex]) as you should easily justify. Let's evaluate the first set:
    [itex]x-1 \geq 0[/itex] and [itex]x+3 \geq 0[/itex]
    implies that
    [itex]x \geq 1[/itex] and [itex]x \geq -3[/itex]
    which is the set [itex]\{x: x \geq 1\}[/itex]. Remember that x must satisfy both inequalities when using "and".
    The second set evaluates to [itex]\{x: x \leq -3\}[/itex], so we have the set [itex]\{x: x \geq 1[/itex] or [itex]x \leq -3\}[/itex], or written another way [itex]\{x: x \geq 1\} \cup \{x: x \leq -3\}[/itex].
    This is just the purely algebraic way. Choose whichever way you feel most comfortable with. :smile:
     
    Last edited: Oct 28, 2005
  7. Oct 28, 2005 #6
    My favorite word: VISUALIZE.

    For generic inequality
    [tex]f(x) > g(x)[/tex]:
    [tex]h(x) = f(x) - g(x)[/tex].
    Find the intervals where a graph [tex]y=h(x)[/tex] is above the x-axis (you'll have to find/estimate the roots of the [tex]y=h(x)[/tex] and points where [tex]h(x)[/tex] is undetermined).

    Try
    [tex]\frac{x+2}{x}\leq \frac{1}{2-x}[/tex]
    Could you post your answer?
     
    Last edited: Oct 28, 2005
  8. Oct 28, 2005 #7
    gosh, that looks confusing!

    i was just taught to regard the inequality as a quadratic, make it equal to 0, draw the graph and solve it from there.

    that may be what ^^ was saying though...
     
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